Let A be the largest positive integer that divides all the numbers of the form $3^k + 4^k + 5^k$, and B be the largest positive integer that divides all the numbers of the form $4^k + 3(4^k) + 4^{k + 2}$, where k is any positive integer. Then (A + B) equals

To find the largest positive integer that divides ALL numbers of this form (by altering k), we need to find the Greatest Common Divisor (GCD) of the first few values.

When k = 1: $3^1 + 4^1 + 5^1 = 3 + 4 + 5 = 12$

When k = 2: $3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50$

When k = 3: $3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$

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GCD(12, 50) = 2 (since 12 = 2² × 3 and 50 = 2 × 5²)

GCD(12, 216) = 2 (since 216 = 2³ × 3³)

Here's the key insight:

$3^k$ is always odd (odd number raised to any power stays odd)

$4^k$ is always even (even number raised to any power stays even)

$5^k$ is always odd (odd number raised to any power stays odd)

Therefore: $3^k + 4^k + 5^k$ = odd + even + odd = even

Since every number of this form is even, they're all divisible by 2. And from our examples, we know that 2 is the largest even number which is a divisor.

Therefore: A = 2

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$4^k + 3(4^k) + 4^{k+2}$

Let us factor out $4^k$:

$= 4^k + 3 \cdot 4^k + 4^k \cdot 4^2$

$= 4^k(1 + 3 + 16)$

$= 4^k \times 20$

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When k = 1: $4^1 \times 20 = 80$

When k = 2: $4^2 \times 20 = 320$

When k = 3: $4^3 \times 20 = 1280$

All these numbers have the form $4^k \times 20$. The GCD will be determined by the smallest power of 4, which is $4^1 = 4$.

Therefore: GCD = $4^1 \times 20 = 80$

Therefore: B = 80

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$A + B = 2 + 80 = 82$

When finding the largest divisor of infinitely many numbers, calculate the GCD of the first few terms. The pattern often becomes clear quickly.